A rigid rod of length is oriented exactly along the radius from Jupiter's centre

PHYSICS · Multi-part · Fermi University Physics · Relativity & Gravitation · Newtonian gravitation as a field theory · Newtonian gravitation as a field theory

The Geometry of Gravity, Ch The tidal field: what free fall cannot remove

Problem

A rigid rod of length ℓ=10.0 m\ell = 10.0\ \mathrm{m} is oriented exactly along the radius from Jupiter's centre, at Io's orbital distance r=4.22×108 mr = 4.22\times10^{8}\ \mathrm{m}. Treat Jupiter as a point mass MJ=1.898×1027 kgM_{\rm J} = 1.898\times10^{27}\ \mathrm{kg} and use G=6.674×10−11 N m2 kg−2G = 6.674\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}. Ignore the rod's own gravity and any non-gravitational forces. ((a)) Write Eij\mathcal{E}_{ij} in radially aligned axes as a diagonal matrix with numerical entries in s−2\mathrm{s^{-2}}, and confirm that its trace vanishes. ((b)) Find the tidal acceleration −EijSj-\mathcal{E}_{ij}S^j between the two ends of the rod.

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